package cn.icatw.leetcode.editor.cn;
//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
//
// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
//
//
//
// 示例 1：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCCED"
//输出：true
//
//
// 示例 2：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"SEE"
//输出：true
//
//
// 示例 3：
//
//
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
//"ABCB"
//输出：false
//
//
//
//
// 提示：
//
//
// m == board.length
// n = board[i].length
// 1 <= m, n <= 6
// 1 <= word.length <= 15
// board 和 word 仅由大小写英文字母组成
//
//
//
//
// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
//
// Related Topics 数组 字符串 回溯 矩阵 👍 1861 👎 0


//Java：单词搜索
public class T79_WordSearch {
    public static void main(String[] args) {
        Solution solution = new T79_WordSearch().new Solution();
        // TO TEST
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public boolean exist(char[][] board, String word) {
            int m = board.length;
            int n = board[0].length;

            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (dfs(board, word, i, j, 0)) {
                        return true;
                    }
                }
            }
            return false;
        }

        private boolean dfs(char[][] board, String word, int i, int j, int k) {
            if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != word.charAt(k)) {
                return false;
            }
            if (k == word.length() - 1) {
                return true;
            }
            char temp = board[i][j];
            board[i][j] = '/';
            boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i, j - 1, k + 1);
            board[i][j] = temp;
            return res;

        }


    }
//leetcode submit region end(Prohibit modification and deletion)

}
